Manually converting a string to an integer in Java

Question

I'm having string consisting of a sequence of digits (e.g. "1234"). How to return the String as an int without using Java's library functions like Integer.parseInt?

public class StringToInteger {
  public static void main(String [] args){
    int i = myStringToInteger("123");
    System.out.println("String decoded to number " + i);
  }

  public int myStringToInteger(String str){
      /* ... */
  }
}

Answer 1

And what is wrong with this?

int i = Integer.parseInt(str);

EDIT :

If you really need to do the conversion by hand, try this:

public static int myStringToInteger(String str) {
    int answer = 0, factor = 1;
    for (int i = str.length()-1; i >= 0; i--) {
        answer += (str.charAt(i) - '0') * factor;
        factor *= 10;
    }
    return answer;
}

The above will work fine for positive integers - assuming that the string only contains numbers, otherwise the result is undefined. If the number is negative you'll have to do a little checking first, but I'll leave that as an exercise for the reader.

Answer 2

A very neat way to convert strings to int- sees contract as an arraylist input and outputs either null(if not convertable) or a valid value.

package <package>;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.stream.IntStream;

public class SO79073232 {
    public static void main(String ...args) {
        ArrayList<String> strs = new ArrayList<>(Arrays.asList("1","2","200","",null, "two"));
        strs.stream().map(SO79073232::toInt).forEach(System.out::println);
    }

    private static int getInt(char c) {
        if (Character.isDigit(c)) {
            return (int) (c - '0');
        } else {
            throw new ArithmeticException(String.format("char %c is not a digit", c));
        }
    }

    private static Integer toInt(String s) {
        try {
            return IntStream.range(0, s.length()).map(i -> (int) (Math.pow(10, s.length() - i - 1)) * (int) getInt(s.charAt(i))).sum();
        } catch (Exception e) {
           return null;
        }
    }
}

it outputs:

1
2
200
0
null
null

The beauty is how it leverages 'streams' and map/reduce of Java lambda programming.

Answer 3

    //Take one positive or negative number
    String str="-90997865";

    //Conver String into Character Array
    char arr[]=str.toCharArray();

    int no=0,asci=0,res=0;

    for(int i=0;i<arr.length;i++)
    {
       //If First Character == negative then skip iteration and i++
       if(arr[i]=='-' && i==0)
       {
           i++;
       }

           asci=(int)arr[i]; //Find Ascii value of each Character 
           no=asci-48; //Now Substract the Ascii value of 0 i.e 48  from asci
           res=res*10+no; //Conversion for final number
    }

    //If first Character is negative then result also negative
    if(arr[0]=='-')
    {
        res=-res;
    }

    System.out.println(res);

Answer 4

Works for Positive and Negative String Using TDD

//Solution

public int convert(String string) {
    int number = 0;
    boolean isNegative = false;
    int i = 0;
    if (string.charAt(0) == '-') {
        isNegative = true;
        i++;
    }

    for (int j = i; j < string.length(); j++) {
        int value = string.charAt(j) - '0';
        number *= 10;
        number += value;
    }
    if (isNegative) {
        number = -number;
    }

    return number;
}

//Testcases

public class StringtoIntTest {
private StringtoInt stringtoInt;


@Before
public void setUp() throws Exception {
stringtoInt = new StringtoInt();
}

@Test
public void testStringtoInt() {
    int excepted = stringtoInt.convert("123456");
    assertEquals(123456,excepted);
}

@Test
public void testStringtoIntWithNegative() {
    int excepted = stringtoInt.convert("-123456");
    assertEquals(-123456,excepted);
}

}

Answer 5

public class ConvertInteger {

public static int convertToInt(String numString){
    int answer = 0, factor = 1;

    for (int i = numString.length()-1; i >= 0; i--) {
        answer += (numString.charAt(i) - '0') *factor;
        factor *=10;
    }
    return answer;
}

public static void main(String[] args) {

    System.out.println(convertToInt("789"));
}

}

Answer 6

This is the Complete program with all conditions positive, negative without using library

import java.util.Scanner;
public class StringToInt {
 public static void main(String args[]) {
  String inputString;
  Scanner s = new Scanner(System.in);
  inputString = s.nextLine();

  if (!inputString.matches("([+-]?([0-9]*[.])?[0-9]+)")) {
   System.out.println("error!!!");
  } else {
   Double result2 = getNumber(inputString);
   System.out.println("result = " + result2);
  }

 }
 public static Double getNumber(String number) {
  Double result = 0.0;
  Double beforeDecimal = 0.0;
  Double afterDecimal = 0.0;
  Double afterDecimalCount = 0.0;
  int signBit = 1;
  boolean flag = false;

  int count = number.length();
  if (number.charAt(0) == '-') {
   signBit = -1;
   flag = true;
  } else if (number.charAt(0) == '+') {
   flag = true;
  }
  for (int i = 0; i < count; i++) {
   if (flag && i == 0) {
    continue;

   }
   if (afterDecimalCount == 0.0) {
    if (number.charAt(i) - '.' == 0) {
     afterDecimalCount++;
    } else {
     beforeDecimal = beforeDecimal * 10 + (number.charAt(i) - '0');
    }

   } else {
    afterDecimal = afterDecimal * 10 + number.charAt(i) - ('0');
    afterDecimalCount = afterDecimalCount * 10;
   }
  }
  if (afterDecimalCount != 0.0) {
   afterDecimal = afterDecimal / afterDecimalCount;
   result = beforeDecimal + afterDecimal;
  } else {
   result = beforeDecimal;
  }

  return result * signBit;
 }
}

Answer 7

Given the right hint, I think most people with a high school education can solve this own their own. Every one knows 134 = 100x1 + 10x3 + 1x4

The key part most people miss, is that if you do something like this in Java

 System.out.println('0'*1);//48

it will pick the decimal representation of character 0 in ascii chart and multiply it by 1.

In ascii table character 0 has a decimal representation of 48. So the above line will print 48. So if you do something like '1'-'0' That is same as 49-48. Since in ascii chart, characters 0-9 are continuous, so you can take any char from 0 to 9 and subtract 0 to get its integer value. Once you have the integer value for a character, then converting the whole string to int is straight forward.

Here is another one solution to the problem

String a = "-12512";
char[] chars = a.toCharArray();
boolean isNegative = (chars[0] == '-');
if (isNegative) {
    chars[0] = '0';
}

int multiplier = 1;
int total = 0;

for (int i = chars.length - 1; i >= 0; i--) {
    total = total + ((chars[i] - '0') * multiplier);
    multiplier = multiplier * 10;
}

if (isNegative) {
    total = total * -1;
}

Answer 8

Using Java 8 you can do the following:

public static int convert(String strNum)
{
   int result =strNum.chars().reduce(0, (a, b)->10*a +b-'0');
}

1. Convert srtNum to char
2. for each char (represented as 'b') -> 'b' -'0' will give the relative number
3. sum all in a (initial value is 0) (each time we perform an opertaion on a char do -> a=a*10

Answer 9

Maybe this way will be a little bit faster:

public static int convertStringToInt(String num) {
     int result = 0;

     for (char c: num.toCharArray()) {
        c -= 48;
        if (c <= 9) {
            result = (result << 3) + (result << 1) + c;
        } else return -1;
    }
    return result;
}

Answer 10

public static int convertToInt(String input){
        char[] ch=input.toCharArray();
        int result=0;
        for(char c : ch){
            result=(result*10)+((int)c-(int)'0');
        }
        return result;
    }

Answer 11

public int myStringToInteger(String str) throws NumberFormatException 
{
    int decimalRadix = 10; //10 is the radix of the decimal system

    if (str == null) {
        throw new NumberFormatException("null");
    }

    int finalResult = 0;
    boolean isNegative = false;
    int index = 0, strLength = str.length();

    if (strLength > 0) {
        if (str.charAt(0) == '-') {
            isNegative = true;
            index++;
        } 

        while (index < strLength) {

            if((Character.digit(str.charAt(index), decimalRadix)) != -1){   
                finalResult *= decimalRadix;
                finalResult += (str.charAt(index) - '0');
            } else throw new NumberFormatException("for input string " + str);

            index++;
        }

    } else {
        throw new NumberFormatException("Empty numeric string");
    }

    if(isNegative){
        if(index > 1)
            return -finalResult;
        else
            throw new NumberFormatException("Only got -");
    }

    return finalResult;
}

Outcome: 1) For the input "34567" the final result would be: 34567 2) For the input "-4567" the final result would be: -4567 3) For the input "-" the final result would be: java.lang.NumberFormatException: Only got - 4) For the input "12ab45" the final result would be: java.lang.NumberFormatException: for input string 12ab45

Answer 12

Alternate approach to the answer already posted here. You can traverse the string from the front and build the number

 public static void stringtoint(String s){      
    boolean isNegative=false;
    int number =0;      
    if (s.charAt(0)=='-') {
        isNegative=true;            
    }else{
        number = number* 10 + s.charAt(0)-'0';
    }

    for (int i = 1; i < s.length(); i++) {

        number = number*10 + s.charAt(i)-'0';           
    }
    if(isNegative){
        number = 0-number;
    }
    System.out.println(number);
}

Answer 13

Use long instead of int in this case. You need to check for overflows.

public static int StringtoNumber(String s) throws Exception{
    if (s == null || s.length() == 0)
        return 0;
    while(s.charAt(0) == ' '){
        s = s.substring(1);
    }
    boolean isNegative = s.charAt(0) == '-';
    if (s.charAt(0) == '-' || (s.charAt(0) == '+')){
        s = s.substring(1);
    }

    long result = 0l;
    for (int i = 0; i < s.length(); i++){
        int value = s.charAt(i) - '0';
        if (value >= 0 && value <= 9){
            if (!isNegative && 10 * result + value > Integer.MAX_VALUE ){
                throw new Exception();
            }else if (isNegative && -1 * 10 * result - value < Integer.MIN_VALUE){
                throw new Exception();
            }
            result = 10 * result + value;
        }else if (s.charAt(i) != ' '){
            return (int)result;
        }
    }
    return isNegative ? -1 * (int)result : (int)result;
}

Answer 14

You can do like this: from the string, create an array of characters for each element, keep the index saved, and multiply its ASCII value by the power of the actual reverse index. Sum the partial factors and you get it.

There is only a small cast to use Math.pow (since it returns a double), but you can avoid it by creating your own power function.

public static int StringToInt(String str){
    int res = 0;
    char [] chars = str.toCharArray();
    System.out.println(str.length());
    for (int i = str.length()-1, j=0; i>=0; i--, j++){
        int temp = chars[j]-48;
        int power = (int) Math.pow(10, i);
        res += temp*power;
        System.out.println(res);
    }
    return res;
}

Answer 15

Make use of the fact that Java uses char and int in the same way. Basically, do char - '0' to get the int value of the char.

public class StringToInteger {
    public static void main(String[] args) {
        int i = myStringToInteger("123");
        System.out.println("String decoded to number " + i);
    }

    public static int myStringToInteger(String str) {
        int sum = 0;
        char[] array = str.toCharArray();
        int j = 0;
        for(int i = str.length() - 1 ; i >= 0 ; i--){
            sum += Math.pow(10, j)*(array[i]-'0');
            j++;
        }
        return sum;
    }
}

Answer 16

Use this:

static int parseInt(String str) {
    char[] ch = str.trim().toCharArray();
    int len = ch.length;
    int value = 0;
    for (int i=0, j=(len-1); i<len; i++,j--) {
        int c = ch[i];
        if (c < 48 || c > 57) {
            throw new NumberFormatException("Not a number: "+str);
        }
        int n = c - 48;
        n *= Math.pow(10, j);
        value += n;
    }
    return value;
}

And by the way, you can handle the special case of negative integers, otherwise it will throw exception NumberFormatException.

Answer 17

If the standard libraries are disallowed, there are many approaches to solving this problem. One way to think about this is as a recursive function:

1. If n is less than 10, just convert it to the one-character string holding its digit. For example, 3 becomes "3".
2. If n is greater than 10, then use division and modulus to get the last digit of n and the number formed by excluding the last digit. Recursively get a string for the first digits, then append the appropriate character for the last digit. For example, if n is 137, you'd recursively compute "13" and tack on "7" to get "137".

You will need logic to special-case 0 and negative numbers, but otherwise this can be done fairly simply.

Since I suspect that this may be homework (and know for a fact that at some schools it is), I'll leave the actual conversion as an exercise to the reader. :-)

Hope this helps!